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CHANGING FORMS

CHANGING FORMS. GENERAL to Transformational. To General Form – Complete the squar. Remember: the general form of a quadratic function is y = ax 2 + bx +c The transformational form is The standard form is y = a(x-h) 2 + k To GET general form: FOIL and solve for y.

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CHANGING FORMS

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  1. CHANGING FORMS GENERAL to Transformational

  2. To General Form – Complete the squar • Remember: the general form of a quadratic function is y = ax2 + bx +c • The transformational form is • The standard form is y = a(x-h)2 + k • To GET general form: FOIL and solve for y. But how do we go FROM general form?

  3. To General Form – Complete the squar Let’s look at an example: What is the vertex of the quadratic function y = 2x2 + 12x -4? We know that in transformational form the coefficient of ‘x’ is 1 So STEP 1 is to divide every term by ‘a’ STEP 2 is to move the non-x term over.

  4. Completing the Squar The right-hand side of the equation needs to be a perfect square: (x –h)2 So far we have: x2 +6x This looks like: x2 x x x x x x

  5. Completing the Squar x2 x x x x x x To make this into a perfect square we could… x x x x2 x x x Oops…

  6. Completing the Squar x2 x x x x x x To make this into a perfect square we could… Almost…we need to complete the square! x2 x x x x x x We were missing nine 1x1 squares.

  7. Completing the Squar So the PERFECT SQUARE is … BUT…how did we get this? ---------- X + 3 ----------- ---------- X + 3 ----------- x2 x x x Added it to the right hand side. This means we must add it to both sides! x x We took HALF the coefficient of x and squared it. x STEP 3: take half the coefficient of x, square it and add it to both sides (x+3)2

  8. Completing the Square So the PERFECT SQUARE is … BUT…how did we get this? ---------- X + 3 ----------- Step 4: Factor out the 1/a term on the left hand side ---------- X + 3 ----------- x2 x x x x x We need the left hand side to have brackets. x Ta Da! (x+3)2

  9. Completing the square We started with the quadratic in general form y = 2x2 + 12x -4. The same function is in transformational form. So what is the range of y = 2x2 + 12x -4 ? Since the equivalent equation shows us a vertex of (-3,-22) with no reflection, the range is

  10. Practice going FROM general form 1. What is the vertex of the function y = 2x2 - 8x +2? 2. What is the range of the function y = -x2 -5x +1? 3. Put the equation y = 0.5x2 – 3x - 1 into transformational form.

  11. Practice going FROM general form 1. What is the vertex of the function y = 2x2 - 8x +2? Vertex: (2,-6)

  12. Practice going FROM general form 2. What is the range of the function y = -x2 -5x +1? Y-value of the vertex Since there’s a reflection

  13. Practice going FROM general form 3. Put the equation y = 0.5x2 – 3x - 1 into transformational form.

  14. A shortcut…eventually If only there was a way to avoid having to complete the square every time to get the vertex!! Maybe there is… Let’s put the following equations into transformational form. y = 3x2 + 12x - 6 y = ax2 + bx + c STEP 1 is to divide every term by ‘a’ STEP 2 is to move the non-x term over.

  15. A shortcut…eventually STEP 3: take half the coefficient of x, square it and add it to both sides Step 4: Factor out the 1/a term on the left hand side Who cares?

  16. Here’s the shortcut We now see that ANY general equation can be written as We haven’t finished this yet but we already see that the x-value of the vertex is X = -b 2a This is VERY important. For example: What is the axis of symmetry of the function y = 2x2 – 16x +3? To complete the square on this takes time. But

  17. Finding y Example 2: What is the range of the graph of the function y = -x2 – 4x+ 7? We know that the x coordinate of the vertex is But how do I find y? You glade the x- value! You “plug in it, plug it in” The range is NOTE: The vertex of ANY quadratic occurs at x = -b/2a and the max or min value is: f(-b/2a)

  18. Changing forms (quickly) Ex. Graph the parabola defined by the function: f(x) = 3x2 + 12x - 9 If f(x) = 3x2 + 12x – 9, then the vertex is We know that transformational form is best for graphing. Is it still necessary? f(-2) = 3(-2)2 +12(-2)-9 f(-2) = 3*4-24-9 f(-2) = -21 NO! Vertex (-2,-21)

  19. Changing forms (quickly) So f(x) = 3x2 + 12x -9 can now be written: If only we knew the ‘a’ value… Since the VS = 3 we can graph from the vertex: Over 1 up 3 Over 2 up 12 Over 3 up 27

  20. Changing Forms (quickly) What we know: f(x) = 3x2 + 12x – 9 Axis of symmetry Range (and Domain) y-intercept Max/min value But we don’t know the x-intercepts!

  21. Finding our Roots When dealing with quadratics, we will be asked to solve quadratic equations, find the roots orfind the zeros of quadratic equations or find the x-intercepts of the parabola. These all mean: solve for x. Ex. Find the x-intercepts of the graph of the function f(x) = x2 - 2x -15 But we can change this into transformational form. Since there are 2 x terms (an x2 and an x) we cannot undo what’s being done. So, in this form (for now) we’re stuck! f(x) = x2 - 2x -15 y = x2 - 2x -15 y+15 = x2 - 2x y +15 +1 = x2 - 2x +1 y +16 = (x-1)2

  22. Finding our Roots f(x) = x2 - 2x -15 y +16 = (x-1)2 We know that for any x-intercepts: f(x) =0 y +16 = (x-1)2 0+16 = (x-1)2 To undo what’s being done to ‘x’ we need to take the square root of both sides. 16 = (x-1)2 So the 2 x-intercepts are (5,0) and (-3,0) Remember: (-4)(-4) =16 Now, this is actually 2 equations in one

  23. Finding our Roots We’ve found the roots of a quadratic function but it involved us going back to transformational form. Is there a shortcut? When we completed the square on the general form we ended up with this Who cares? We Do If we set y= 0 we can come up with a formula to find the roots of a quadratic function. We’ll call it the “Quadratic Root Formula”

  24. Finding our Roots Now, on the left-hand side we need to add fractions, so we need a common denominator.

  25. Finding our Roots So if we go back to our original example: Ex. Find the x-intercepts of the graph of the function f(x) = x2 - 2x -15 First set one side equal to 0 (since y = 0 for x-intercepts) 0 = x2 - 2x -15 Now plug in a, b and c into the quadratic root formula TA DA!!!!! So the 2 x-intercepts are (5,0) and (-3,0)

  26. Yet another method of finding roots • We have solved quadratic equations by • putting it in transformational form and undoing what’s being done • Making one side 0 and using the quadratic root formula Sometimes we can also factor. This uses the zero property. If (r)(s) = 0 then EITHER ‘r’ must equal 0 or ‘s’ must equal 0. This only works when the product is 0. So if we can get the quadratic function in the form (x-r)(x-s)=0 then we know that either x-r = 0 or x –s has to be zero. This is called FACTORING

  27. Factoring to find the roots Ex. Find the x-intercepts of the graph of the function f(x) = x2 - 2x -15 Answer when the numbers are added Again, f(x) = 0 0 = x2 -2x -15 We need two numbers whose product is -15 and whose sum is -2 -5 3 ___ x ___ = -15 ___ + ___ = -2 Answer when the numbers are multiplied -5 3 So x2 -2x-15=0 (x-5)(x+3)=0 So the 2 x-intercepts are (5,0) and (-3,0) Either (x-5) = 0 Or (x+3) = 0 x = 5 x = -3

  28. Factoring When factoring a quadratic equation where ‘a’ = 1, find two numbers that multiply to give ‘c’ and add to give ‘b’. By the way: 24 = x2 +10x 24 +25 = x2 +10x +25 49 = (x+5)2 +/- 7 = x + 5 x = -5 +7 or x = -5 -7 x = 2 or x = -12 Ex. Solve for x by factoring ___ x ___ = -24 ___ + ____ = 10 12 -2 12 -2 or x =(-10 +14)/2 =2 x = (-10-14)/2 = -12

  29. Practice 1. Solve the following equations by factoring. a) 0= x2 + 2x +1 c) 0= x2 + 2x -24 b) 0= x2 + 5x +4 d) 0= x2 -25 2. Find the x- and y- intercepts of the following quadratics. a) f(x) = 2x2 +3x + 1 c) f(x) = x2 -5x -14 d) y= 2(x-4)2 - 32 b) f(x) = 3x2 -7x + 2 Answers 1a. x = -1 b. x = -4 or -1 c. x = -6 or 4 d. x = -5 or +5 2a. (-1,0) (-0.5, 0) (0,1) b. (1/3,0) (2,0) (0, 2) c. (-2,0) (7,0) (0, -14) d. (0,0) (8,0)

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